sequence of rationals with subsequences converging to every real number I need an example of a sequence of rationals with subsequences converging to every real number but I can't think of any.
I think the fact that Q is dense in R comes into play but I'm confused. 
 A: If we can quote the fact that the set of rationals is countable, then we know that all the rationals can be listed as a sequence $r_0,r_1,r_2,\dots$. Then the result follows from the fact the rationals are dense in $\mathbb{R}$. 
A: First go from $-1$ to $1$ at intervals of $1$.  Then go from $-2$ to $2$ at intervals of $1/2$, then from $-3$ to $3$ at intervals of $1/3$, and so on.  These repeated sweeps will cover the neighborhood of any fixed real number (once they are large enough to reach it) at finer and finer intervals.  A subsequence converging to $x\in\mathbb{R}$ can be formed by choosing the closest element to $x$ in each sweep.
A: Let $q_n$ be an enumeration of the rationals. Then construct the sequence $q_1, q_1,q_2, q_1,q_2, q_3, q_1,q_2, q_3,q_4, \cdots$, that is, write out the $q_n$ from $1$ to $k$, then from $1$ to $k+1$, etc, etc.
Now suppose $x \in \mathbb{R}$. Then we know that there is a sequence of rationals $q_{n_k} \to x$. Then it should be clear that we can find $q_{n_1}$ in the above sequence, then $q_{n_2}$, etc, etc.
Note: I realize that the above is overkill, but I think it makes the answer clear with little thought?
